I'm trying to deserialize a Json string usign Json.NET. I have a basic understanding about it but I've been trying for days and can't make it work, here's my code. (Please ignore any unecessary/extra code, have in mind I was trying multiple solutions based on what I found after researching).

Cannot deserialize the current JSON array (e.g. [1,2,3]) into type 'Twitter.TrendParser+RootObject' because the type requires a JSON object (e.g. {"name":"value"}) to deserialize correctly.

To fix this error either change the JSON to a JSON object (e.g. {"name":"value"}) or change the deserialized type to an array or a type that implements a collection interface (e.g. ICollection, IList) like List that can be deserialized from a JSON array. JsonArrayAttribute can also be added to the type to force it to deserialize from a JSON array.

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As i said before i have a basic understanding of the Json format, and i've been trying using different types of classes to deserialize with the same result, any tip or any idea about what the problem may be?. I am also stuck with .NET 2.0 and not 3rd party libraries aside from json.NET.

Thank you for the attention.

EDIT *

Apparently even if i try declaring List value, the variable "value" won't be created in the actual context, any way to fix that.

Tried this, it was my fault because i didn't mention it but it says that Values doesn't exist in the actual context, tried initializing it in the constructor and got nothing either. Still tried inspecting it and somehow it doesn't have any value after that. Any idea why?
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Edward NewgateAug 31 '12 at 16:06

Error: values The name 'values' does not exist in the actual context. I had to translate it as my compiler is in another lanugage but it translates like that.
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Edward NewgateAug 31 '12 at 16:26

That usually means that you are trying to use the variable values later on in your code. i.e values == null I'd stick a breakpoint after the DeserializeObject call and see if you have anything. It does not sound like JSON.NET is throwing the error.
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James SouthAug 31 '12 at 23:16

Done that, apparently it doesn't recognize the variable no matter how i declare it. Maybe it has something to do with the compiler but not sure, i was taking a look at the assembly code and couldn't figure the error either... Weird.
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Edward NewgateSep 2 '12 at 0:10